Is there more than one possible reduced row echelon form of a matrix?

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I'm trying to find the rref(A) where A = \begin{bmatrix}2&1&4&-4&11\\1&2&3&1&8\\1&1&2&-1&6\end{bmatrix}

I used row operations and came out with the following: \begin{bmatrix}1&0&0&-17&13\\0&1&0&8&-6\\0&0&1&-6&8\end{bmatrix}

However the Linear Algebra Toolkit came out with this: \begin{bmatrix}1&0&0&-3&3\\0&1&0&2&1\\0&0&1&0&1\end{bmatrix}

I read in other questions here that it isn't possible to have two different possible rrefs and I'm unclear as to what I've done incorrectly as the rows satisfy the requirements of a rref matrix. Can anyone identify what I've done incorrectly? I can submit my exact row operations if required. Thanks.

EDIT: Here's a list of my operations: $R_1-R_3,R_2-R_3, R_3-R_2, R_1-2R_2, R_3*R_2, R_2-R_3, R_1-2R_2$

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1 Answer

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The reduced row echelon form is unique. The are many paths to this form, but the destination is unique.

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Let's look at the example problem. Use augmented reduction and clear the columns sequentially.

Column 1$$ % E \left[ \begin{array}{rcc} \frac{1}{2} & 0 & 0 \\ -\frac{1}{2} & 1 & 0 \\ -\frac{1}{2} & 0 & 1 \\ \end{array} \right] % A \left[ \begin{array}{cccrc|ccc} 2 & 1 & 4 & -4 & 11 & 1 & 0 & 0 \\ 1 & 2 & 3 & 1 & 8 & 0 & 1 & 0 \\ 1 & 1 & 2 & -1 & 6 & 0 & 0 & 1 \\ \end{array} \right] % = % \left[ \begin{array}{cccrc|rcc} \boxed{1} & \frac{1}{2} & 2 & -2 & \frac{11}{2} & \frac{1}{2} & 0 & 0 \\ 0 & \frac{3}{2} & 1 & 3 & \frac{5}{2} & -\frac{1}{2} & 1 & 0 \\ 0 & \frac{1}{2} & 0 & 1 & \frac{1}{2} & -\frac{1}{2} & 0 & 1 \\ \end{array} \right] % $$

Column 2$$ % E \left[ \begin{array}{crc} 1 & -\frac{1}{3} & 0 \\ 0 & \frac{2}{3} & 0 \\ 0 & -\frac{1}{3} & 1 \\ \end{array} \right] % A \left[ \begin{array}{cccrc|rcc} \boxed{1} & \frac{1}{2} & 2 & -2 & \frac{11}{2} & \frac{1}{2} & 0 & 0 \\ 0 & \frac{3}{2} & 1 & 3 & \frac{5}{2} & -\frac{1}{2} & 1 & 0 \\ 0 & \frac{1}{2} & 0 & 1 & \frac{1}{2} & -\frac{1}{2} & 0 & 1 \\ \end{array} \right] % = % \left[ \begin{array}{ccrrc|rrc} \boxed{1} & 0 & \frac{5}{3} & -3 & \frac{14}{3} & \frac{2}{3} & -\frac{1}{3} & 0 \\ 0 & \boxed{1} & \frac{2}{3} & 2 & \frac{5}{3} & -\frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & -\frac{1}{3} & 0 & -\frac{1}{3} & -\frac{1}{3} & -\frac{1}{3} & 1 \end{array} \right] % $$

Column 3$$ % E \left[ \begin{array}{ccr} 1 & 0 & 5 \\ 0 & 1 & 2 \\ 0 & 0 & -3 \\ \end{array} \right] % A \left[ \begin{array}{ccrrc|rrc} \boxed{1} & 0 & \frac{5}{3} & -3 & \frac{14}{3} & \frac{2}{3} & -\frac{1}{3} & 0 \\ 0 & \boxed{1} & \frac{2}{3} & 2 & \frac{5}{3} & -\frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & -\frac{1}{3} & 0 & -\frac{1}{3} & -\frac{1}{3} & -\frac{1}{3} & 1 \end{array} \right] % = % \left[ \begin{array}{cccrc|rrr} \boxed{1} & 0 & 0 & -3 & 3 & -1 & -2 & 5 \\ 0 & \boxed{1} & 0 & 2 & 1 & -1 & 0 & 2 \\ 0 & 0 & \boxed{1} & 0 & 1 & 1 & 1 & -3 \\ \end{array} \right] % $$

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